The floor function $\lfloor x\rfloor$ and $((x))$ are intimately linked in computing Dedekind sums - which led me the quest below.
My current interest lies in this:
QUESTION. Assume $a, b, c$ are positive real numbers. Is this inequality true? $$\lfloor 3a\rfloor+\lfloor 3b\rfloor+\lfloor 3c\rfloor\geq 2\lfloor a\rfloor+2\lfloor b\rfloor+2\lfloor c\rfloor+\lfloor a+b+c\rfloor.$$
As mentioned by @MichaelRozenberg, we may assume $\{a,b,c\}\subset[0,1)$.
So write $a=\frac i3+\alpha$, $b=\frac j3+\beta$, $c=\frac k3+\gamma$, where $i\in\{0,1,2\}$, $j\in\{0,1,2\}$, $k\in\{0,1,2\}$, and $\alpha\in[0,\frac13)$, $\beta\in[0,\frac13)$, $\gamma\in[0,\frac13)$. The LHS of your inequality is $$\lfloor i+3\alpha\rfloor +\lfloor j+3\beta\rfloor+\lfloor k+3\gamma\rfloor=i+j+k,$$ while the RHS of your inequality is $$\left\lfloor \frac{i+j+k}3 +\alpha+\beta+\gamma\right \rfloor.$$ Now $0\le\alpha+\beta+\gamma<1$ so $i+j+k= \lfloor i+j+k+\alpha+\beta+\gamma\rfloor$. The inequality follows from the fact that $i+j+k\ge \frac{i+j+k}3$.
\left and \right modifiers to make \lfloor and \rfloor automagically adjust their size to the height of an expression between them. For example, \lfloor\frac k3\rfloor results in $$\lfloor\frac k3\rfloor$$ whilst \left\lfloor\frac k3\right\rfloor looks like $$\left\lfloor\frac k3\right\rfloor$$
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Commented
May 4, 2023 at 18:52
\frac{\sqrt{1\times\frac 1x}}3 + \left.\frac xy\right|_{x=1}^5 makes $$\frac{\sqrt{1\times\frac 1x}}3 + \left.\frac xy\right|_{x=1}^5$$ The left-dot and right-bar delimit the x/y ratio, so the bar fits that ratio's height only, despite the left addend being higher.
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